Logarithmic Differentiation

[AlexDest]

Holy fuck these problems I was given to work on (had all of class time to do it in class, now it's homework).

y = x*sqrt(x² + 1) / (x + 1)^2/3

What I got for the derivative (taking natural log of both sides):

y' = ((x^3 + 9x² + 5x + 3) / (3x(x² + 1)(x + 1)) * (x*sqrt(x² + 1) / (x + 1)^2/3)

Our teacher doesn't want us to anymore with that so that would be a good enough answer to her (lol high school).

Here's more:

y = cubedroot(x*(x - 2) / x^2 + 1)

y' = (-2(3x² - x + 1) / 3x(x^3 - x - 2)) * (cubedroot(x*(x - 2) / x^2 + 1))

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y^5 = sqrt((x+1)^5 / (x+2)^10)

y' = (15 / 2(x² + 3x + 2)) * (sqrt((x+1)^5 / (x+2)^10)) / 5)

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y = cubedroot((x)(x+1)(x-2) / (x² + 1)(2x + 3))

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y^4/5 = sqrt(sinxcosx) / 1 + 2 ln x

y' = (-xcosxsinx - 2cosxsinx - 2xsinxcosx / 2xsinxcosx + 4sinxcosx) * (5 * (sqrt(sinxcosx) / 1 + 2 ln x) / 4)

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sqrt(y) = x^5 * arctanx / (3 - 2x)*cubedroot(x)

[dag12]

Re: Logarithmic Differentiation

I won't do these since they're just grindy algebra, but wolframalpha.com should help.

[smartdude1212]

Re: Logarithmic Differentiation

Wolfram is useful, but in moderation. Plus they've started limiting the solutions you can see unless you subscribe.

I'll write these out on paper and scan for you to see, and I can explain everything in detail after if you would like. I find it easier than trying to blunder through ridiculous algebra symbols on a forum.

[Netjet!]

Re: Logarithmic Differentiation

Just got home, but I'll pull out my pen and paper and see if I can give you a hand.

[smartdude1212]

Re: Logarithmic Differentiation

Voila!

Hopefully I didn't screw anything up, and hopefully I copied down your questions properly. I didn't do one of them oops w/e.
I know, don't post answers, yadda yadda, but this is just answer or die stuff here.

[Ohaider]

Re: Logarithmic Differentiation

damn +1 smartdude

[AlexDest]

Re: Logarithmic Differentiation

Voila!

Hopefully I didn't screw anything up, and hopefully I copied down your questions properly. I didn't do one of them oops w/e.
I know, don't post answers, yadda yadda, but this is just answer or die stuff here.

i taste blood atm

i'm doing copious amounts of unnecessary factoring lmfao

[smartdude1212]

Re: Logarithmic Differentiation

When it comes to ridiculous derivatives, factoring is pointless unless you absolutely need to simplify (which typically only happens when graphing because then you need the second derivative, etc.).

You'll notice the only simplifying I do is to make simple expressions like x(x-2) easier to work with (rather than split that up into ln(x) + ln(x-2) I would much rather deal with ln(x^2-2x) anyway, since ln derivatives are simple).

Also, for what I labelled #4, you have:
y' = (-xcosxsinx - 2cosxsinx - 2xsinxcosx / 2xsinxcosx + 4sinxcosx) * (5 * (sqrt(sinxcosx) / 1 + 2 ln x) / 4).

Let's face it, those sin(x)*cos(x) terms look disgusting, but they're easy to get rid of as long as you remember that sin(2x) = 2*sin(x)*cos(x) (and thus sin(x)*cos(x) = sin(2x)/2). Hell, simplifying like that may even impress your teacher.

The alternative: split them up as ln(sin(x))+ln(cos(x)) (but who wants to do that these days ).

And my answer for the one I neglected to do:

y' = y{1/(3x) + 1/[3*(x + 1)] + 1/[3*(x - 2)] - (2x)/[3*(x^2 + 1)] - 2/[3*(2x + 3)]}, where y = cubedroot((x)(x+1)(x-2) / (x² + 1)(2x + 3)) (when I learned this I never had to replace what y was simply because it was unnecessary writing)